1133 Splitting A Linked List (25 分)（链表）

2022-03-02 23:02:46 阅读：111 来源： 互联网

标签：node 25 10 int list next 链表 1133 id

Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 105) which is the total number of nodes, and a positive K (<=1000). The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer in [-105, 105], and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218

Sample Output:

33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

题目大意：

给一个链表和K，遍历链表后将<0的结点先输出，再将0～k区间的结点输出，最后输出>k的结点

分析：

将结点用list[10000]保存，list为node类型，node中保存结点的值value和它的next地址。list的下标就是结点的地址。将<0、0～k、>k三部分的结点地址分别保存在v[0]、v[1]、v[2]中，最后将vector中的值依次输出即可～

原文链接：https://blog.csdn.net/liuchuo/article/details/78037305

题解

#include <bits/stdc++.h>

using namespace std;
const int maxn=100000;
struct Node
{
    int address,key,next;
}node[maxn];
int vis[maxn];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,s,id,k;
    cin>>s>>n>>k;
    for(int i=0; i<n; i++)
    {
        cin>>id;
        node[id].address=id;
        cin>>node[id].key>>node[id].next;
    }
    vector<int> v[3];
    for(int i=s;i!=-1;i=node[i].next){
        if(node[i].key<0){
            v[0].push_back(i);
        }else if(node[i].key>k){
            v[2].push_back(i);
        }else v[1].push_back(i);
    }
    int flag=0;
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < v[i].size(); j++) {
            if (flag == 0) {
                printf("%05d %d ", v[i][j], node[v[i][j]].key);
                flag = 1;
            } else {
                printf("%05d\n%05d %d ", v[i][j], v[i][j], node[v[i][j]].key);
            }
        }
    }
    printf("-1");
    return 0;
}

标签：node,25,10,int,list,next,链表,1133,id
来源： https://www.cnblogs.com/moonlight1999/p/15957602.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。

ICode9